Melody

Nice work Yesoka,

I got the same answer via a different method.

Use pythagoras theorem twice to set up 2 equations

Sove simultaneously first to get x and then to get the height

It is not there.

Hi Jurimagic,

A squared function is always symmetric.   The axis of symmetry (for a concave up or down one is   \(x=\frac{-b}{2a}\)

you can get the roots by adding and minusing   \(\frac{\sqrt{b^2-4ac}}{2a}\)  to the axis of symmetry

Think about the most simple one.  \(y=x^2\)

\((+x)^2=(-x)^2 \)       so it must be symmetricall about y=0

Cubic functions are never symetrical about a line.   

eg:  concider y=x^3

\((+x)^3\ne(-x)^3\)     so it is fefinitelyt not symmetrical about a line,

You need to do lots of work with these and slowly the logic of it all will seep into your brain.

There are not any such integers.

But the easiest way to solve it is to let the integers be   x, x+1, x+2, x+3, x+4, x+5, x+6

now do the equation bit.   X equals a decimal, not an integer

This question has been asked many tims recently. 
One of the others has proper answers.

I will unlock the original posts for you. I only answered one of them.

Please explain what you do not understand about my answer on the original post 

This has been answered in a nearby post.

\( |x| = -\dfrac 1 2 x + 4\\ |x| = \frac {-x+8}{2}\\ \text{square both sides}\\ x^2=\frac{x^2-16x+64}{4}\\ 4x^2=x^2-16x+64\\ 3x^2+16x-64=0\\ x=\frac{-16\pm\sqrt{256+768}}{6}\\ x=\frac{-16\pm32}{6}\\ x=\frac{-8\pm16}{3}\\ x=-8\quad or \quad x=\frac{8}{3}\\ \text{BUT these answers need to be checked for validity.} \)

Check x=-8

RHS= 4+4=8=LHS     works

check x=8/3

RHS= -8/6 +4 = -4/3 +4 = (-4+12)/3 = 8/3 = LHS   also works

LaTex

 |x| = -\dfrac 1 2 x + 4\\
 |x| = \frac {-x+8}{2}\\
\text{square both sides}\\
x^2=\frac{x^2-16x+64}{4}\\
4x^2=x^2-16x+64\\
3x^2+16x-64=0\\
x=\frac{-16\pm\sqrt{256+768}}{6}\\
x=\frac{-16\pm32}{6}\\
x=\frac{-8\pm16}{3}\\
x=-8\quad or \quad x=\frac{8}{3}\\
\text{BUT these answers need to be checked for validity.}

g^-1(7)   does not exist.

When questions get a bit old they get automatically locked. So that is probably why.  Look for the lock symbol.

If you want an old question unlocked so that you can answer it, just send a moderator a privae message, with a link, and ask them to unlock it for you. 

Don't forget the please :) 

Note:  it will not stay unlocked for ever.